Question

`\int\limits^5_1 { (lnR)/( R^(2) ) \, dR`

Answer 1

Integrate by parts, setting


`\begin{matrix}u=\ln R&&\mathrm dv=(\mathrm dR)/(R^2)\\\mathrm du=\frac{\mathrm dR}R&&v=-\frac1R\end{matrix}`

So the integral is


`\displaystyle\int_1^5(\ln R)/(R^2)\,\mathrm dR=-\frac{\ln R}R\bigg|_(R=1)^(R=5)+\int_1^5(\mathrm dR)/(R^2)`

`\displaystyle\int_1^5(\ln R)/(R^2)\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_(R=1)^(R=5)`

`\displaystyle\int_1^5(\ln R)/(R^2)\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)`

`\displaystyle\int_1^5(\ln R)/(R^2)\,\mathrm dR=\frac15(4-\ln5)`